Ex 2.4, 4 If two zeroes of the polynomial x4 – 6×3 – 26×2 + 138x – 35 are 2 ± √3, find other zeroes. Introduction Factorising 6 2 is a factor of 6 3 is a factor of 6 So, 2 × 3 is also a factor of 6 We will use the same in our question Hence, (x − 2 − √3) × (x − 2 + √3) is also a factor. ( (x − 2) − √3) × ( (x − 2) + √3) is also a factor. (x − 2)2 − (√3)2 x2 + 22 − 4x − 3 x2 + 4 − 4x − 3 x2 − 4x + 1 ∴ x2 − 4x + 1 is a factor of p(x) Now by dividing p(x) by (x2 − 4x + 1) We can find out other factors Now, we find zeroes of x2 − 2x − 35 x2 – 2x + 25 = 0 We find the zeroes using Splitting the middle term method x2 − 7x + 5x − 35 = 0 x (x −7) + 5 (x − 7) = 0 (x + 5) (x − 7) = 0 So, x = –5 and x = 7 are zeros Therefore, the zeroes of p(x) are 2 + √ , 2 − √ , −5 and 7

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Ex 2.4, 4 If two zeroes of the polynomial x4 – 6×3 – 26×2 + 138x – 35 are 2 ± √3, find other zeroes. Introduction Factorising 6 2 is a factor of 6 3 is a factor of 6 So, 2 × 3 is also a factor of 6 We will use the same in our question Hence, (x − 2 − √3) × (x − 2 + √3) is also a factor. ( (x − 2) − √3) × ( (x − 2) + √3) is also a factor. (x − 2)2 − (√3)2 x2 + 22 − 4x − 3 x2 + 4 − 4x − 3 x2 − 4x + 1 ∴ x2 − 4x + 1 is a factor of p(x) Now by dividing p(x) by (x2 − 4x + 1) We can find out other factors Now, we find zeroes of x2 − 2x − 35 x2 – 2x + 25 = 0 We find the zeroes using Splitting the middle term method x2 − 7x + 5x − 35 = 0 x (x −7) + 5 (x − 7) = 0 (x + 5) (x − 7) = 0 So, x = –5 and x = 7 are zeros Therefore, the zeroes of p(x) are 2 + √ , 2 − √ , −5 and 7

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